Class 9 Science Chapter 9 Force and Law of Motion NCERT Solutions
Question Answers Page 118
1.Which of the following has more inertia:
(a) a rubber ball and a stone of the same size?
(b) a bicycle and a train?
(c) a five rupees coin and a one-rupee coin?
Answer :
Inertia is the measure of mass of the body.The greater is the mass of a body, greater is its inertia and vice versa.
a) Mass of the stone is more than the mass of rubber ball for same size.Hence, inertia of the stone is greater than that of rubber ball.
b) Mass of train is more than mass of a bicycle. Hence , inertia of train is more than that of bicycle.
c) Mass of Rs 5 coin is more than that of Rs 1 coin. Hence, inertia of Rs 5 coin is more than that of Rs 1 coin.
2. In the following example, try to identify the number of times the velocity of the ball changes: “A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”. Also identify the agent supplying the force in each case.
The velocity of ball changes four times.
First when a football player kicks to another player.Agent supplying the force is first player.
Second when the player kicks the football to the goal keeper.Agent supplying the force is second player.
Third when the goalkeeper stops the ball.Agent supplying the force is goalkeeper.
Fourth when the goalkeeper kicks the ball towards a player of his own team.Agent supplying the force is goalkeeper.
3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer:
When a tree is shaken vigorously, its leaves fall down because tree move to and fro but its leaves remain at rest due to their inertia and hence detach from the tree and fall down.
4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Answer:
In a moving bus, passengers move with bus due to inertia of motion.As the driver applies break, the bus comes to rest.But the passengers tries to maintain his state of motion.As a result a forward force is exerted on him.
Passengers tend to fall backwards when the bus accelerates from rest.This is because when the bus accelerates, the inertia of the passengers tends to oppose the forward motion of the bus.Hence, the passengers tends to fall backwards when the bus accelerates forward.
Page 126, 127
1. If action is always equal to the reaction, explain how a horse can pull a cart.
Answer:
A horse pushes the ground with foot in the backward direction. According to Newton’s Third law of motion, a reaction force is exerted by the earth on the horse in the forward direction. As a result the cart moves forward.
2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Answer:
It is difficult for a fireman to hold a hose, which ejects large amount of water at a high velocity , then a reaction force is exerted on him by ejecting water in backward direction.This is because of Newton’s Third Law of motion.As a result of the backward force, the stability of fireman decreases.
3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 ms–1 . Calculate the initial recoil velocity of the rifle.
Answer:
Mass of the rifle , m1 = 4 kg
Mass of the bullet , m2 = 50 g = 0.05 g
Recoil velocity of the rifle = v1
Bullet is fired with an initial velocity= v2 = 35 m/s
Initially bullet is at rest.Thus its initial velocity , v=0
Total initial momentum of the rifle and bullet system = (m1+m2)v =0
Total momentum of the rifle and bullet system after firing = 4 (v1) + 0.05 × 35
= 4 v1 + 1.75
According to law of conservation of momentum
Total momentum after firing = Total momentum before firing
4V1 + 1.75 =0
V1= – 0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.
4. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms–1 and 1 ms–1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms–1 . Determine the velocity of the second object.
Answer:
Mass of one of the object , m1 = 100g
Mass of the other object , m2 = 200 g
Velocity of m1 before collision , v1 = 2 m/s
Velocity of m2 before collision , v2 = 1 m/s
Velocity of m1 after collision, v3 = 1.67 m/s
Velocity of m2 after collision , v4 =
According to the law of conservation of mass
Total momentum before collision= Total momentum after collision
m1v1 = m2v2
0.1 × 2 + 0.2 × 1 = 0.1 × 1.67 + 0.2 × v4
v4 = 1.165 m/s
The velocity of the second object becomes 1.165 m/s after the collision.
Exercise 128, 129
1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Answer:
Yes, even when an objects experience a net zero external unbalanced force, it is possible that object travels with non-zero velocity.This is possible only when the object moves with a constant velocity in particular direction.Then there is no net unbalanced force applied on the body.The object will keep moving with a non-zero velocity.To change the state of motion, a net non-zero unbalanced external force must be applied on the object.
2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer:
When a carpet is beaten with stick , the carpet is set into motion.But the dust particles tend to remain in state of rest due to inertia.Hence, dust particles comes out of the carpet.
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:
When a moving bus suddenly stops, the luggage on its roof tends to continue in the state of motion due to inertia. Hence the luggage fall down from the roof of the bus. Similarly, when a bus suddenly starts, the luggage on the roof of the bus tends to continue in the state of rest and hence fall down from the roof of the bus. Thus, to avoid the falling of the luggage, it is tied with a rope on the roof of a bus.
4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.
Answer:
The ball slows to a stop because there is a force on the ball opposing the motion.It is the frictional force which always acts in the direction opposite to the direction of motion.
5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
Answer:
Initial velocity, u = 0
Distance covered by stone = 400 m
Time = 20 s
According to Second equation of motion
S =ut + ½ at2
400 = 0 × 20 + ½ × a ×(20 )2
a= 2 m/s2
Force =mass × acceleration
F= 7000 × 2 = 14000 N
6. A stone of 1 kg is thrown with a velocity of 20 ms–1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Answer:
Initial velocity of the stone , u = 20 m/s
Final velocity of the stone , v = 0 m/s
Distance covered by stone = 50 m
According to Third Law of motion
v2 =u2 + 2aS
(0)2 = (20)2 + 2 × a × 50
a= -4 m/s2
Mass of the stone = 1 kg
From Newton’s second law of motion
F= Ma
F = 1 × (-4)
F= -4 N
7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force and (b) the acceleration of the train.
Answer:
a) Force exerted by the engine , F = 40000 N
Frictional force offered by the track = Ff = 5000 N
Accelerating force = Fa = F -Ff = 40000- 5000 = 35000 N
b) Acceleration of the train = a
The engine exerts a force of 40000 N on all the five wagons.
Net accelerating force on the wagon = Fa = 35000 N
Mass of the wagons = mass of wagon × number of wagons
m = 2000 × 5
m= 10000 Kg
Total mass (M) = m + 8000 = 10000+ 8000 = 18000 kg
Fa = Ma
a = Fa /M
a= 35000/18000
a= 1.944 m/s2
c) Mass of 4 wagons except wagon 1 = 2000 × 4= 8000 Kg
Acceleration of the train = 1.944 m/s2
The force of wagon 1 on wagon 2 = ma = 8000 × 1.944 = 15552 N
8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms–2?
Answer:
Mass of automobile vehicle , m=1500 kg
Final velocity , v = 0 m/s
Acceleration of the automobile , a= -1.7 m/s2
F=Ma
F = 1500 × (-1.7)
F = -2550 N
The force between the automobile and the road is -2550 N , in the direction opposite to the motion of the automobile.
9. What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2 (b) mv2 (c) ½ mv2 (d) mv
Answer: (d) Momentum =mv
10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer:
A force of 200 N is applied in the forward direction.Thus from Newton’s third law of motion , an equal amount of force will act in opposite direction.This opposite force is the frictional force exerted on the cabinets.Hence a frictional force of 200 N is exerted on the cabinets.
11. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m s-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Answer:
Mass of one of the object , m1 = 1.5 Kg
Mass of the other object , m2 = 1.5 Kg
Velocity of m1 before collision, v1 = 2.5 m/s
Velocity of m2 moving in opposite direction after collision = v2 = -2.5 m/s
Total mass of combined object = m1 + m2
Velocity of combined object = v
According to law of conservation of momentum
Total momentum before collision= Total momentum after collision
m1v1 + m2v2 = (m1+ m2) v
1.5(2.5) +1.5(-2.5) = (1.5+1.5) v
v=0
The velocity of the combined objects after collision is 0 m/s.
Page 129
12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer:
The truck has a large mass.Therefore the static force between the truck and the road is also very high.To move a car, one has to apply a force more than the static force.Therefore when someone pushes the truck and truck does not move, then it can be said that the applied force in one direction is cancelled out by the frictional force of equal amount acting in the opposite direction.
13. A hockey ball of mass 200 g travelling at 10 ms–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms–1. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer:
Mass of hockey ball , m = 200 g =0.2 kg
Hockey ball travels with velocity , v1 = 10 m/s
Initial momentum = mv1
Hockey ball travels in the opposite direction with velocity , v2 = -5 m/s
Final momentum = mv2
Change in momentum = mv1 -mv2 = 0.2 [ 10- (-5)]= 3 Kg m s-1
Hence the change in momentum of the hockey ball is 3 Kg m s-1
14. A bullet of mass 10 g travelling horizontally with a velocity of 150 ms–1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Answer:
velocity of Bullet = 150 m/s
When bullet enters the block , its initial velocity, u = 150 m/s
Final velocity , v= 0 m/s
Time taken to come to rest= 0.03 s
v= u +at
0 = 150 + (a × 0.03)
a= -5000 m/s2
v2 = u2 -2aS
(0)2 = (150)2 – 2 × (-5000) × S
S= 2.25 m
Distance of penetration of bullet into the block is 2.25 m.
F=ma
mass of the bullet ,m= 10 g = 0.01 Kg
Acceleration of the bullet , a= 5000 m/s2
F=ma
F= 0.01 × 5000
F= 50 N
15. An object of mass 1 kg travelling in a straight line with a velocity of 10 ms–1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Answer:
Mass of the object , m1= 1 kg
Velocity of the object before collision, v1 = 10 m/s
Mass of stationary wooden block, m2 = 5 Kg
Velocity of wooden block after collision , v2 = 0 m/s
Total momentum before collision= m1v1+ m2v2
Total momentum= 1(10) + 5(0) = 10 kg m/s
After collision object and wooden block stick together
Total mass of combined system= m1 + m2
Velocity of the combined object = v
According to law of conservation of momentum
Total momentum before collision= Total momentum after collision
m1v1 +m2v2 = (m1+m2)v
1(10) + 5(0) = (1+5) v
v= 5/3 m/s
v= 1.67 m/s
16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms–1 to 8 ms–1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer:
Initial velocity of the object , u = 5 m/s
Final velocity of the object , v = 8 m/s
Mass of the object = m= 100 Kg
Time taken by the object to accelerate , t = 6 s
Initial momentum = mu= 100 × 5 = 500 kg ms-1
Final momentum = mv = 100 × 8 = 800 kg ms-1
F= mv-mu / t
F = m(v-u) /t
F= 50 N
Initial momentum= mu = 500 kg ms-1
Final momentum = mv = 800 kg ms-1
Force exerted on the object is 50 N.
17. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Answer:
As per the law of conservation of momentum. Momentum of the car and insect system before collision =Momentum of the car and insect system after collision. Hence, the change in momentum of the car and insect is zero.
The velocity of the insect changes to a great amount as a result of collision between insect and car and velocity of insect changes to a great amount. On the side, car continues moving with constant velocity. Hence, the Kiran’s suggestion that the insect suffers great change in momentum as compare to the Car is correct. The momentum of the insect after collision becomes very high because the car is moving at a high speed. So, the momentum gained by the insect is equal to the momentum lost by the Car.
Akhtar’s statement is correct because the mass of the Car is very large as compare to the mass of the insect.
Rahul’s statement is also correct as both the Car and insect experienced equal force caused by the Newton’s action reaction law. But, he also made a wrong statement as the system suffers change in momentum because the momentum before the collision is equal to momentum after the collision.
18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 ms–2 .
Answer:
Mass of the dumb bell = 10 kg
Distance covered by the dumb bell = s=80 cm= 0.8 m
Acceleration in the downward direction, a= 10 m/s2
Initial velocity of the dumb bell , u =o
Final velocity of the dumb bell=
According to the third equation of motion:
v2 = u2 + 2as
v2 = 0+ 2(10) (0.8)
v= 4 m/s
Momentum with which the dumb bell hits the floor is mv
Momentum= 10 × 4 = 40 Kg ms-1
Additional Exercises 130
A1. The following is the distance-time table of an object in motion:
Answer:
| Time in seconds | Distance in meters |
| 0 | 0 |
| 1 | 1 |
| 2 | 8 |
| 3 | 27 |
| 4 | 64 |
| 5 | 125 |
| 6 | 216 |
| 7 | 343 |
(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
(b) What do you infer about the forces acting on the object?
Answer:
a) There is an unequal change of distance in an equal interval of time.Thus the given object is having non-uniform acceleration.Since the velocity of object increases with time , the acceleration is increasing.
b)According to Newton’s Second law of motion, the force acting on an object is directly proportional to the acceleration produced in the object.The increasing acceleration of the given object indicates that the force acting on the object is also increasing.
A2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms-2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
Answer:
Mass of the motor cycle = 1200 Kg
Acceleration produced by the car, when it is pushed by the third person , a= 0.2 m/s2
Force applied by the third person =F
Force = mass × acceleration
F = 1200 × 0.2 = 240 N
Thus the third person applies a force of magnitude 240 N.
A3. A hammer of mass 500 g, moving at 50 m s-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Answer:
Mass of the motor car, m= 1200 kg
Initial velocity of the motor car , u= 90 km/hr= 25 m/s
Final velocity of the motor car , v = 18 km/hr = 5 m/s
Time taken , t = 4 s
v= u + at
5 = 25 + a(4)
a= -5 m/s2
Change in momentum = mv-mu = m(v-u)
Change in momentum= 1200 (5-25) = -24000 Kg m/s
Force = mass × acceleration
F= 1200 × (-5)
F= -6000 N
Acceleration of the motor car= -5 m/s2
Change in momentum of the motor car = -24000 Kg m/s
The force required to decrease the velocity is 6000 N.
A4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Answer:
Mass of the motor cycle , m= 1200 Kg
Initial velocity of the car , u= 90 Km/hr =25 m/s
Final velocity of the car, v= 18 Km/hr= 5 m/s
Time taken ,t= 4 s
v= u + at
5 = 25 + a(4)
a=-5 m/s2
Change of momentum = m(v-u)
Change of momentum= 1200(5-25) = -24000 Kg ms-1
Force = Mass × Acceleration
Force = 1200 × (-5) = -6000 N
Acceleration of the motor car = -5 m/s2
Change in momentum of the motor car = -24000 Kg ms-1
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